HLT362v Statistics-Frequency distribution

Exercise 32

Calculating t-tests for Paired (Dependent) Samples

Answer:

Exercise 31

1. The given data meets all the assumptions for an independent samples t-test. The validation for this is that:

• Independent observations: Each case represents a different statistical unit (Controlled and supported employment).
• Normality: The dependent variable follows a normal distribution in the population. The test for normality is as shown in the table below:

Table 1: Normality Test

Only the test of the Shapiro-Wilk is focused on since the cases are less than 2000. Since P > 0.05, the null hypothesis is cannot be rejected and it can be established that the data is derived from a normal distribution.

• Homogeneity: since the sample sizes are equal, there was no need to conduct the homogeneity test. The assumption is only tested when the sample sizes are (sharply) unequal (Mayers, 2013).

                                                                                 Figure 1: Frequency distribution

From figure 1, it is evident that the shape of the distribution is bell-shaped, thus the variable follows a normal distribution.

From table 1, the outcome of the test of normality for Shapiro-Wilk was a statistics of 0.935 with a p-value of 0.194. Hence the conclusion that the data is derived from a normal distribution since the p-value is greater than 0.05.

3. Table 2: Groups descriptive statistics

The means of the wages received per week of the control group is $128.40  $43.03 while the mean of the wages received per week of the supported employment is $232.70  $65.33.

4. Table 3: Independent Samples Test

From the Levene’s test, since the significance is greater than 0.05. Thus, the tests of equal variances assumed holds. Therefore, the independent sample t-test value is -4.217.

5. From the test on equal variances assumed it is seen that the significance is less than 0.05. Thus, we choose to not accept the null hypothesis since the t-test is significant statistically and conclude that the population means are not equal.

6. From table 3 above, it is evident that the precise probability of obtaining a t-test value which is either as extreme or as close to the one that was really perceived with the assumption that the null hypothesis is true is 0.1%.

7. From table 3 above, the mean difference is -$104.30.  Thus, the change between the control group and the supported employment group is -$104.30. Therefore, it can be concluded that the group which earned most money post-treatment is the supported employment group.

8. The independent sample t-test was conducted with an aim to compare wage payment between supported employment group and control group. There was a significant difference in the averages for the control group (M=$128.4, SD=$43.03) and the supported employment group (M=$232.7, SD=$65.33); t(18)=-4.2, p=0.001.

9. The outcomes suggest that the type of employment has an effect on the amount of wages one receives. Thus, the supported employment vocational rehabilitation impacts the wages earned by disabled veterans positively. The program can, therefore, be deemed to be beneficial.

10. The sample size can be stated to be adequate in detecting the difference that is significant between the two groups. The rationale of this argument can be supported by the results showing the contrast between the average of the earned money, the t value, the p-value, and the groups’ confidence interval.

Exercise 32

Calculating t-tests for Paired (Dependent) Samples

Based on figure 2 and 3 above, it is evident that the shapes of the distributions are bell-shaped. Therefore, the two variables follow a normal distribution.

Table 4: Tests of Normality

From table 4 above, the outcomes of the Shapiro-Wilk test of normality was a statistics of 0.953 with a p-value of 0.705 for MPI Affective Distress Baseline and a statistics of 0.912 with a p-value equal to 0.292 for MPI Affective Distress Post Tx. Hence the conclusion that the two variables have data coming from a normal distribution since the p-values are greater than 0.05.

3. Table 5: Paired Samples Statistics

The average of the MPI Affective Distress Baseline is 3.03  1.66 while the average of the MPI Affective Distress Post Tx is 2.04  0.98.

4. Table 6: Paired Samples Test

The paired sample t-test value as seen in table 6 is 2.865.

5. From table 6 above, the significance is 0.019. Since the p-value is less than 0.019, we choose to accept the null hypothesis. Thus, the t-test is significant at the alpha level where it is equal to 0.05.

7. Evidently, the distress scores deteriorated over time. On average, the MPI Affective Distress Baseline was higher than the MPI Affective Distress Post TX (95% CI [0.21, 1.77]) as seen by the mean difference of 0.99.

8. The paired samples t-test was carried out so as to relate the different distress scores before rehabilitation on emotional distress. There was a difference that was significant in the MPI Affective Distress Baseline (M=3.03, SD=1.66) and MPI Affective Distress Post Tx (M=2.04, SD=0.983); t(2.865)=, p=0.019.

9. The outcomes, therefore, suggest that the rehabilitation of emotional distress impacts positively on the stress levels of an individual.

10. The study design involves the use of a low sample size. Thus, the derived results cannot be generalized without using caution.

Reference:

Mayers, A. (2013). Introduction to Statistics and SPSS in Psychology. Pearson Higher Ed.

Ross, A., & Willson, V. L. (2017). Paired Samples T-Test. In Basic and Advanced Statistical Tests (pp. 17-19). SensePublishers, Rotterdam.